$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is:

$A$ parallel plate capacitor is made of two square parallel plates of area $A$,and separated by a distance $d << \sqrt{A}$. The capacitor is connected to a battery with potential $V$ and allowed to fully charge. The battery is then disconnected. $A$ square metal conducting slab also with area $A$ but thickness $d/2$ is then fully inserted between the plates,so that it is always parallel to the plates. How much work has been done on the metal slab by an external agent while it is being inserted?

$A$ parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers,parallel to its plates,each with thickness $\delta = \frac{d}{N}$. The dielectric constant of the $m^{\text{th}}$ layer is $K_m = K(1 + \frac{m}{N})$. For a very large $N (> 10^3)$,the capacitance $C$ is $\alpha \left( \frac{K \varepsilon_0 A}{d \ln 2} \right)$. The value of $\alpha$ will be. . . . . . . .
[$\varepsilon_0$ is the permittivity of free space]

$A$ capacitor is half-filled with a dielectric $(K=2)$ as shown in Figure $A$. If the same capacitor is to be filled with the same dielectric as shown in Figure $B$, what would be the thickness $t$ of the dielectric so that the capacitor still has the same capacity?

$A$ parallel plate air capacitor has capacity $C$ farad, potential $V$ volt, and energy $E$ joule. When the gap between the plates is completely filled with a dielectric of dielectric constant $K$, what happens to the potential $V$ and energy $E$?

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become: